Finding all groups of order $7$ up to isomorphism? I'm learning group theory but I didn't learn any concepts of building groups.
I know that there exists the identity group $\{e\}$, and the group with 2 elements: $\{e,a\}$.
If I try to create a group with 3 elements, let's say: $\{e,a,b\}$ then we would have:
$ea = a, eb = b, aa = ?$, and what about $ab$?
Am I supposed to try this for the $7$ elements?
What does the statement ``up to isomorphism" mean? In particular, how many groups of order $7$ are there up to isomorphism?
I'm really confused, and my book says nothing about it.
 A: Comment:  This answer was put up before @egreg's above, but that answer should be the one used at the level of the op,  go to that one!
Groups are "Unique" up to isomorphism.  Isomorphism in this sense means that if we relabeled the elements of one group in the right way, we'd get the exact other group.   For instance, there's two groups of order 4:  The cyclic group,  and the Klein 4 group.  The Klein 4 group can be represented as $\mathbb Z_2 \times \mathbb Z_2$ with ordered pair addition,  or as $\{e,a,b,ab\}$,  with everything commuting and being order 2.  The two groups are the exact same up to isomorphsim, in that if we labeled $e=(0,0),a=(1,0),b=(0,1),ab=(1,1)$,  all the group operations on the left and right hand sides of those equality would be the same.  (that function that does the labeling is called the isomorphism as well)
There's a cyclic group of every order.   For primes, cyclic groups are the only group (A simple consequence of Lagrange's theorem that the order of an element divides the order of the group),  and obviously for n=1 there's only 1 group, so that takes care of 1,2,3,5,7.   4 and 6 require more powerful tools to prove you know all of them.
For 4,  the easiest tool is once you've proved that any group of the order of a prime squared must be abelian,  and then with the fundemental theorem of finite abelian groups, you get it must either be $\mathbb Z _4$ or $\mathbb Z _2 \times \mathbb Z_2$.   Doing it without those tools would be annoying, I'm not sure what your teacher wants if you're at the beginning.
For 6,  by the same fundemental theorem of finite abelian groups, there's only 1 abelian group.   To prove there's exactly one nonabelian group, the easiest tool I know of is semidirect products....once you learn semidirect products,  it's an early proof to show that if a group has order $pq$,  $p<q$,  with both $p$ and $q$ prime and $p|(q-1)$,  then there's exactly one nonabelian group of that order.  In this case, the nonabelian group can be represented as $D_6$ (The dihidral group, or symmetries of a triangle)  or $S_3$  (The symmetric group on 3 symbols). In this case, they are isomorphic.
A: When p is prime then there is only one group upto isomorphism,namely cyclic group of order p. So from this we get there exist unique group of order 2,3,5 and 7.
So only non-trivial cases are group of order 4 and 6..Any group of oder $p^2$ where p is a prime is an abelian group.So any  group of order 4 is abelian.It is easy to check that there exist upto isomorphidm two subgroups namely cyclic group $Z_4$ and klien group $V_4$.One can check that upto isomorphism there exist exactly two group of order 6,namely symmetric group $S_3$ and cyclic group $Z_6$.
Note:
For order 4,you can do it from basic definition. You know in any group order 4 there are either elements of order 1,2 or4.If you have any element of order 4,then the group is cyclic.Suppose there does not exist any element of order 4.This implies order of all elements is 2.So this proves the group is abelian and non-cyclic,this group is known as klien 4-group.
There is a theorem which says that a group of order 2p where p is a odd prime upto isomorphism there exist two subgroups $Z_6$   and $D_3$  ,dihedral group of order 6.To prove this you only need sylow thorem.see this..
Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .
A: Groups of order $1$, $2$, $3$, $5$ and $7$ are unique up to isomorphism: for $1$ it's obvious, the others are prime numbers.
Let's tackle the $4$ case; one of the elements is the identity, so the group is $\{1,a,b,c\}$. Every group of even order has an element of order $2$ (a very easy case of Cauchy's theorem). Let's say it's $a$, so $a^2=1$.
Make the Cayley table for the group, where we fill in the information we already have. 
$$
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 \\
b & b\\
c & c\\
\end{array}
$$
For the product $ab$ we have only one choice, namely $c$, because an element cannot appear twice in a row or column. Hence $ac=b$. Similarly, $ba=c$ and $ca=b$. So we have
$$
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c \\
c & c & b \\
\end{array}
$$
Now we have two choices for $c^2$: either $c^2=1$ or $c^2=a$. Let's look at what we have, because once we decide for $c^2$ the other slots can be filled in uniquely:
$$
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & 1 & a \\
c & c & b & a & 1
\end{array}
\qquad\qquad
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & a & 1 \\
c & c & b & 1 & a
\end{array}
$$
The structure on the left is Klein's group, the one on the right is the cyclic group. So we have two groups of order $4$.
For the groups with six elements we need something more advanced, because the Cayley table method is too long if started from nothing. Another application of Cauchy's theorem says that the group has a cyclic subgroup of order $3$. So there are three elements $1,a,b$ such that $a^2=b$, $ab=ba=1$ and $b^2=a$. Let's fill in this information for the set $\{1,a,b,c,d,e\}$. Again we can also assume $c^2=1$.
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 &   &   &   \\
b & b & 1 & a &   &   &   \\
c & c &   &   & 1 &   &   \\
d & d \\
e & e
\end{array}
$$
For $ac$ we can choose either $ac=d$ or $ac=e$; it doesn't matter which one, because at the end exchanging $d$ with $e$ will give an isomorphism; so let's choose $ac=d$ that forces $ad=e$ and $ae=c$
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a &   &   &   \\
c & c &   &   & 1 &   &   \\
d & d \\
e & e
\end{array}
$$
Let's now examine $ca$: again we have either $ca=d$ or $ca=e$, but this time we can't choose one and we need to go with two alternatives:
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a &   &   &   \\
c & c & d &   & 1 &   &   \\
d & d & e \\
e & e & c
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a &   &   &   \\
c & c & e &   & 1 &   &   \\
d & d & c \\
e & e & d
\end{array}
$$
We now have no choice for $bc$: in both cases it must be $e$; similarly, $bd=c$ and also $cb$ has a determined result (and we can also fill in some other slots):
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & d & e & 1 &   &   \\
d & d & e & c &   &   &   \\
e & e & c & d &   &   &
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & e & d & 1 &   &   \\
d & d & c & e &   &   &   \\
e & e & d & c
\end{array}
$$
We're almost done. Observe that interchanging $a$ and $b$ is irrelevant, because both are generators of the (normal) subgroup with three elements. So we can arbitrarily choose $cd=a$ in both cases and we can complete the tables. Indeed, if in the table on the left we choose $dc=b$, we'd have $a(dc)=ab=1$, but $(ad)c=ec=a$ against associativity. Similarly for the table on the right.
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & d & e & 1 & a & b \\
d & d & e & c & a & b & 1 \\
e & e & c & d & b & 1 & a
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & e & d & 1 & a & b \\
d & d & c & e & a & b & 1 \\
e & e & d & c & b & 1 & a
\end{array}
$$
So we have a maximum of two groups of order six. But we know two non isomorphic one: the cyclic group and $S_3$. So we're done.
A: Up to isomorphism, there is only one group of prime order $p$.
Any group of prime order $p$ isomorphic to $C_p$: the cyclic group of order $p$.
